First slide

Motion of centre of mass

Question

A man weighing 80 kg is standing on a trolley weighting 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley along the rails at speed I m/s (w.r.t to trolley) then after 4 s his displacement relative to the ground will be:

Moderate

A

5 m
B

4.8 m
C

3.2 m
D

3.0 m

Solution

As no extemal force acting on the system, the displacement of centre of mass of the system should be zero. Let displacement of the trolley in opposite direction of motion of the man in four seconds is x.
Hence $∆ x_{cm} = 0 = \frac{m ∆ x_{man} + M ∆ x_{trolley}}{m + M}$
$80 [(1 \times 4) - x] + 320 (- x) = 0, x = \frac{80 \times 4}{(320 + 80)} = 0.8 m$
Therefore the net displacement of man = 4 - 0.8 = 3.2 m

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